Julia's Food Booth

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Date Added: 08 / 23 / 2011
Category: Business and Marketing
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Julia's Food Booth

View Full Essay Assignment 3:   Julia’s Food Booth
Misti Brinson
Dr. Matt Johnson
Quantitative Methods - MAT

Assignment #3:   Julia’s Food Booth
  A. Formulate and solve an L.P. model for this case:

X1 = pizza slices
X2 = hot dogs
X3 = B B Q sandwiches

This model is set up for the first home game.
Next, you would maximize Z = $0.75 x1 + 1.05 x2 + 1.35 x3 which is subject to the following:
$0.75x1 + 0.45x2 + 0.90x3 <= 1,500 thus leading to:
24x1 + 16x2 + 25x3 <= 55,296 in^2 of even space:   thus you have x 1>=*x2 + x3 giving you
  X2
_____   >= 2.0                 so x1, x2, x3 >=0      
  X3

All of which is displayed in the graph below. (Note: The oven space needed for a slice of pizza is determined by dividing the total space required by the slice of pizza, 14 x 14 = 196 in^2 by 8 or 24 in^2 per slice. The total space available would be the dimensions of the shelf, 36 in. x 48 in. = 1,728 in^2, multiply that by 2, the times before kickoff and halftime, thus the oven will be filled 55,296in^2.

Solution: X1 = 1,250 slices of pizza
X2 = 1,250 hot dogs
      X3 = 0   bbq sandwiches
Z   =   $2,250
| | | | | | | | |
Food items: |   | Pizza | Hot Dogs | Barbecue | | | | |
Profit per item: | 0.75 | 1.05 | 1.35 | | | | |
Constraints: |   |   |   | Available | Usage | Left over | |
Budget ($) | 0.75 | 0.45 | 0.90 |       1,500 |     1,500.00 | 0 | |
Oven space (sq. in.) | 24 | 16 | 25 |     55,296 |   50,000.00 | 5296 | |
Demand | 1 | -1 | -1 | 0 |             -   | 0 | |
Demand | 0 | 1 | -2 | 0 |     1,250.00 | -1250 | |
| | | | | | | | |
Stock | | | | | | | | |
Pizza= | 1250 | slices | | | | | | |
Hot Dogs= | 1250 | hot dogs | | | | | | |
Barbecue= | 0 | sandwiches | | | | | | |
Profit= |   2,250.00 | | | | | | | |

    Julia’s profit before expenses should be $2,250. The lease is $1,000 per game thus giving a profit of...